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Question 1170358: Hi. This is my second question for today.
A bridge has an elliptical arch as a support. The arch has a height of 7 meters and a width at the base of 40 meters. A horizontal roadway is 2 meters above the center of the arch. How far would it be above the arch at 8 meters from the center?
I've been solving this but can't be quite sure whether I'm correct or not. I got 0.6 m.
Thank you so much.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Draw an xy axis. Plot the point A at the origin (0,0). This will be the center of the ellipse. So (h,k) = (0,0).
Then plot points B and C at (-20,0) and (20,0) respectively. The distance from B to C is 40 meters, which is the width of the arch.
Note the length of the semi-major axis is 20 units. The semi-major axis is horizontal, so we'll make a = 20.
Point D will go at (0,7). The semi-minor axis is vertical, so b = 7 represents the length of the semi-minor axis.
With those h,k, a and b values, we get
(x-h)^2/(a^2) + (y-k)^2/(b^2) = 1
(x-0)^2/(20^2) + (y-0)^2/(7^2) = 1
(x^2)/400 + (y^2)/49 = 1
This represents the equation of the ellipse
The upper roadway is represented by a horizontal line through 9 on the y axis. This is because the upper roadway is 2 meters above the highest point D.
Here's the graph of everything discussed so far. Point E wasn't discussed earlier, but I mention it in the next section.
Go back to point A. Move 8 meters either right or left. I'll move 8 meters to the right. Then you move upward to point E as shown in the diagram. Currently the location of point E is (8, p).
What we do is plug x = 8 into the ellipse equation we set up. Then we solve for y
(x^2)/400 + (y^2)/49 = 1
(8^2)/400 + (y^2)/49 = 1
64/400 + (y^2)/49 = 1
4/25 + (y^2)/49 = 1
(y^2)/49 = 1 - 4/25
(y^2)/49 = 25/25 - 4/25
(y^2)/49 = 21/25
sqrt[ (y^2)/49 ] = sqrt(21/25)
y/7 = sqrt(21)/sqrt(25)
y/7 = sqrt(21)/5
y = 7*sqrt(21)/5
y = (7/5)*sqrt(21)
y = 1.4*sqrt(21)
So we can see that E is located at ( 8, 1.4*sqrt(21) ).
The last step will have us subtract the y coordinate of point E from y = 9, which is the height of the upper roadway. This will tell us the vertical distance from E to the upper roadway.
So we get 9 - 1.4*sqrt(21) = 2.58439402706182
Exact Answer: 9 - 1.4*sqrt(21) meters
Approximate Answer: 2.58439402706182 meters
Round the approximate value however you need.
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